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I have no idea how to start!

  1. Sep 2, 2012 #1
    1. A ball launched from ground level lands 2.5 s later on a level field 30 m away from the launch point. Find the magnitude of the initial velocity vector and the angle it is above the horizontal. (Ignore any effects due to air resistance.)



    2. Known Kinematic Equations



    3. I don't know how to start!
     
  2. jcsd
  3. Sep 2, 2012 #2

    trollcast

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    As per the homework section guidelines I can only give hints but have you considered the horizontal and vertical components separately as once you work them out you can work out the resultant and then use a bit of trigonometry to find the angle.
     
  4. Sep 2, 2012 #3
    Yes, however, when I solved for initial velocity using x=v0t+1/2at^2, I got -0.25s. That doesn't sound right.
     
  5. Sep 2, 2012 #4

    trollcast

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    Thats not the way I worked it out,

    Start with the horizontal component as it doesn't change because we are ignoring the effects of air resistance.

    So it travels 30m horizontally in 2.5sec, therefore

    $$V = \frac{distance}{time} $$
    $$V = \frac{30}{2.5} $$
    $$V = 12ms^{-1} $$

    Then consider the vertical component,

    The initial velocity is unknown.
    The acceleration will be -g
    The time taken to reach the apex of the parabola will be half the total time.
    The final velocity at the apex will be 0.

    Can you work out the initial vertical component from that?
     
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